Question

Evaluate the limit:
$\underset{x\to \infty }{lim}{\left\{\frac{x^2+2x+3}{2x^2+x+5}\right\}}^{\frac{3x-2}{3x+2}}$

Answer 1

We have,
$\underset{x\to \infty }{lim}{\left\{\frac{x^2+2x+3}{2x^2+x+5}\right\}}^{\frac{3x-2}{3x+2}}$

$=\underset{x\to \infty }{lim}{\left\{\frac{x^2(1+\frac{2}{x}+\frac{3}{x^2})}{x^2(2+\frac{1}{x}+\frac{5}{x^2})}\right\}}^{\frac{x(3-\frac{2}{x})}{x(3+\frac{2}{x})}}$

$=\underset{x\to \infty }{lim}{\left[\frac{(1+\frac{2}{x}+\frac{3}{x^2})}{(2+\frac{1}{x}+\frac{5}{x^2})}\right]}^{\frac{(3-\frac{2}{x})}{(3+\frac{2}{x})}}$

As $x\to \infty \Rightarrow \frac{1}{x}\to 0$

$={\left[\frac{1+0+0}{2+0+0}\right]}^{\frac{(3-0)}{(3+0)}}$

$=\frac{1}{2}$

Therefore,
$\underset{x\to \infty }{lim}{\left\{\frac{x^2+2x+3}{2x^2+x+5}\right\}}^{\frac{3x-2}{3x+2}}=\frac{1}{2}$